Question

For a first order reaction, rate constants at \(50^\circ\mathrm{C}\) and \(100^\circ\mathrm{C}\) are \(1.5\times10^{-5}\,\mathrm{s^{-1}}\) and \(4.5\times10^{-5}\,\mathrm{s^{-1}}\), respectively. What is the approximate activation energy of the reaction (in \(\mathrm{kJ\,mol^{-1}}\))? \[ (\log3=0.48,\;R=8.3\ \mathrm{J\,mol^{-1}K^{-1}}) \]

• A. \(22\)
• B. \(28\)
• C. \(38\)
• D. \(44\)

Hint:

For two temperatures, \[ \boxed{ \log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R} \left( \frac1{T_1}-\frac1{T_2} \right). } \] Always convert temperatures into Kelvin before substitution.

Answer 1

The Correct Option is A

Step 1: Use the Arrhenius equation. For two temperatures, \[ \log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R} \left( \frac1{T_1}-\frac1{T_2} \right). \] Here, \[ k_1=1.5\times10^{-5}, \qquad k_2=4.5\times10^{-5}, \] so that \[ \frac{k_2}{k_1}=3. \] Also, \[ T_1=323\ \mathrm{K}, \qquad T_2=373\ \mathrm{K}. \]

Step 2: Substitute the values. Using \[ \log3=0.48, \] \[ 0.48 = \frac{E_a}{2.303\times8.3} \left( \frac1{323}-\frac1{373} \right). \] Now, \[ \frac1{323}-\frac1{373} = \frac{50}{323\times373} \approx4.15\times10^{-4}. \] Hence, \[ E_a = \frac{0.48\times2.303\times8.3} {4.15\times10^{-4}} \approx2.2\times10^4\ \mathrm{J\,mol^{-1}}. \]

Step 3: Convert into kJ mol\(^{-1}\). \[ E_a \approx22\ \mathrm{kJ\,mol^{-1}}. \] Hence, \[ \boxed{22\ \mathrm{kJ\,mol^{-1}}.} \] Therefore, the correct option is \(\boxed{(A)}\).